Question

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance $D=2.20m$ from the edge of the table;see Fig.$8-48.$ Bobby compresses the spring $1.10cm,$ but the center of the marble falls $27.0cm$ short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

Answer 1

The distance the marble travels is determined by its initial speed (and the methods of Chapter $4$), and the initial speed is determined (using energy conservation) by the original compression of the spring. We denote h as the height of the table, and x as the horizontal distance to the point where the marble lands.

Then $x=v_{0}t$ and $h=\frac{1}{2}g{t}^2$(since the vertical component of the marble's “launch velocity” is zero). From these we find $x=v_0\sqrt{2h/g}.$. We note from this that the distance to the landing point is directly proportional to the initial speed. We denote $v_{01}$ be the initial speed of the first shot and $D_1=(2.20-0.27)m=1.93m$be the horizontal distance to its landing point; similarly, $v_{02}$ is the initial speed of the second shot and $D=2.20m$ is the horizontal distance to its landing spot. Then

$\frac{v_{02}}{v_{01}}=\frac{D}{D_1}\Rightarrow v_{02}=\frac{D}{D_1}{v}_{01}$

When the spring is compressed an amount $l$ , the elastic potential energy is $\frac{1}{2}k{l}^2$. When the marble leaves the spring its kinetic energy is $\frac{1}{2}m{v}_0^2$. Mechanical energy is conserved: $\frac{1}{2}m{v}_0^2=\frac{1}{2}k{l}^2,$ and we see that the initial speed of the marble is directly proportional to the original compression of the spring. If $l_1$ is the compression for the first shot and $l_2$ is the compression for the second, then $v_{02}=\left(l_2/l_1\right)v_{01}.$ Relating this to the previous result, we obtain

$l_2=\frac{D}{D_1}{l}_1=\left(\frac{2.20m}{1.93m}\right)\left(1.10cm\right)=1.25cm$